Work in the rest frame of the planar shock, with nonzero mass flux $j=\rho u$. Conservation of mass gives $[j]=0$; the tangential and normal components of balance of linear momentum give $[jv]=0$ and $[p+ju]=0$. Thus $[v]=0$. For an ideal gas, specific internal energy is $e=c_{v}T=p/[(\gamma -1)\rho ]$, and specific enthalpy is $e+p/\rho =\gamma p/[(\gamma -1)\rho ]$. Total energy flux is $j[(u^2+v^2)/2+e+p/\rho ]$. Its continuity, with $v$ already continuous, givesPut $c^2=\gamma p/\rho$ and $M=u/c$. Since $p=ju/(\gamma M^2)$, the normal component of balance of linear momentum says $u[1+1/(\gamma M^2)]$ is constant. HenceThe energy equation says $u^2[1/2+1/((\gamma -1)M^2)]$ is constant, givingSet $a=M_{-}^2$, $b=M_{+}^2$, square the first relation and equate to the second. After clearing positive denominators the equation factors asThe first factor is the unchanged state. For a genuine shock the second factor vanishes, yieldingThe physical selection principle is entropy increase in the direction of flow. For $u_{\pm }>0$, the upstream side is $x<0$ and the downstream side is $x>0$. A compressive entropy-admissible shock has $\boxed{M_{-}>1,M_{+}<1}$; interpreting the printed “ahead” and “behind” spatial labels must respect these explicit side assignments. The expansion-shock branch would violate the second law.

Substitution gives the velocity ratioSince $v$ is unchanged while the normal component decreases, $|v|/u$ increases. Thus an oblique flow bends towards the shock plane, equivalently away from its normal. If $v=0$ there is no angular bending; the flow stays normal and simply slows. The tangential continuity, not a change in the tangential speed, causes the deflection.

Long-wave geometry makes vertical acceleration small, so pressure is approximately hydrostatic, $p=p_{\mathrm{atm}}+\rho g(h-z)$. The leading horizontal velocity is nearly depth-independent. Integrating incompressibility through the depth and using the impermeable bed and kinematic free-surface condition gives $\boxed{h_t+(hu{)}_x=0}$. The horizontal Euler equation then gives $\boxed{u_t+u{u}_x+g{h}_x=0}$.

Linearizing about a uniform resting layer gives wave speed $\boxed{c=\sqrt{gh}}$. Since $h=c^2/g$, the nonlinear equations can be written $c_t+u{c}_x+(c/2)u_x=0$ and $u_t+u{u}_x+2c{c}_x=0$. Combining them verifiesLet $c_0=\sqrt{g{h}_0}$ and $V\ge 0$ be the receding wall's speed, so its position is $x=-Vt$ and its fluid velocity condition is $u=-V$. The outward-moving rarefaction has the other invariant fixed at its resting value, $u-2c=-2c_0$. In its varying part, self-similarity gives $x/t=u+c$. ThereforeThe wall-adjacent state has $u=-V$ and $c=c_w=c_0-V/2$. Its right-going characteristic curve speed is $c_0-3V/2$, whereas the undisturbed head travels at $c_0$. The complete solution in the fluid $x>-Vt$ isBoth edges match continuously and the wall velocity condition holds. The assumption $V^2<4g{h}_0$ means $V<2c_0$, giving positive wall depth and placing the fan tail to the right of the wall by speed $c_w>0$. The depth everywhere is $h=c^2/g$. This is a shallow-water rarefaction from a receding wall.

For a vector $a$ described in axes rotating at constant angular velocity $\Omega$, the rotating-axes theorem is $(da/dt{)}_{\mathrm{inertial}}=(da/dt{)}_{\mathrm{rotating}}+\Omega \times a$. Inertial particle velocity is $u+\Omega \times x$. Applying the theorem once more gives inertial acceleration $Du/Dt+2\Omega \times u+\Omega \times (\Omega \times x)$. Thus the rotating momentum equation isIf $g=-\nabla \Phi$ and density is constant, the centrifugal term can be absorbed into the reduced pressure in a rotating fluid:For small disturbances with $\Omega =\Omega e_z$, linearize to $u_t-2\Omega v=-P_x/\rho$, $v_t+2\Omega u=-P_y/\rho$, $w_t=-P_z/\rho$, together with incompressibility. Put $\zeta =v_x-u_y$. Taking the vertical curl gives ${\zeta }_t=2\Omega w_z$. Taking divergence gives ${\nabla }^{2}P=2\rho \Omega \zeta$. Applying the Laplacian to the vertical momentum equation therefore gives ${\partial }_t{\nabla }^{2}w=-2\Omega {\zeta }_z$. Another time derivative yieldsAll three velocity components are retained, even for fields independent of $y$; rotation produces the transverse component.

Set $m=\pi /h$ and seek $w=a{e}^{-i\omega t}W(x)\sin mz$. The rigid-wall conditions are then automatic and the scalar equation givesFor $0<|\omega |<2|\Omega |$, define $q=m\sqrt{4{\Omega }^2/{\omega }^2-1}$. The two possible oscillatory factors are distinguished by energy propagation. With horizontal wavenumber $k$, the dispersion relation is ${\omega }^2=4{\Omega }^2{m}^2/(k^2+m^2)$ and $c_{g,x}=-\omega k/(k^2+m^2)$. An outgoing wave in $x>0$ must have $\omega k<0$, givingFor $|\omega |>2|\Omega |$, define $\kappa =m\sqrt{1-4{\Omega }^2/{\omega }^2}$. Reject the exponentially growing branch, leaving the evanescent responseThe real part is the physical velocity. The radiation or bounded-decay condition is needed in addition to the wavemaker boundary value to pick a unique half-channel response. Steady forcing ($\omega =0$) is a singular limit: with nonzero rotation a steady geostrophic balance and impermeable top/bottom require $w=0$, incompatible with a nonzero imposed sine profile. At the cutoff equality, the bounded outgoing polarization also degenerates. These exceptional cases cannot be obtained by blindly substituting into the two strict-inequality formulas. The propagating sign is the radiation sign for an inertial wave.

For a material volume, mass conservation is $d{\int }_{V(t)}\rho dV/dt=0$. The transport theorem and arbitrariness of that volume give $\boxed{{\rho }_t+\nabla \cdot (\rho u)=0}$. Balance of linear momentum includes the pressure traction and body force:Apply transport, the divergence theorem and the mass equation to obtainFor small perturbations of the uniform resting barotropic state, write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=p_0+c_0^2{\rho }^{\prime }+\cdots$. Linearized mass and momentum areTaking the curl shows that initially vanishing vorticity remains zero, so the acoustic velocity has a potential $u=\nabla \varphi$. Integrating momentum, using the additive freedom of a velocity potential to remove the spatially constant integration term, gives ${\rho }^{\prime }=-({\rho }_0/c_0^2){\varphi }_t$. Continuity now yields $\boxed{{\varphi }_{tt}=c_0^2{\nabla }^2\varphi }$.

For spherical symmetry, set $\varphi =e^{-i\omega t}F(r)$ and $k=\omega /c_0$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }+k^2{r}^{2}F=0$, or $(rF{)}^{\prime \prime }+k^2(rF)=0$. HenceRigid boundaries require radial velocity $F^{\prime }=0$ at both radii. A form automatically satisfying the condition at $a$ isAt $b$, the condition $b(rF{)}^{\prime }-(rF)=0$ becomesTherefore the nonzero oscillatory eigenfrequencies obeyThe undivided sine-cosine form avoids artificial issues from dividing by a cosine. The formal $k=0$ root is the spatially constant zero-frequency potential additive freedom of a velocity potential, not a propagating acoustic oscillation. This is the radial acoustic spectrum between rigid concentric spheres.

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.

Scale lateral distance with $L$, transverse distance with $h$, lateral velocity with $U$, and the normal velocity with $Uh/L$. Balancing lateral pressure force with transverse viscous stress gives pressure scale $\mu UL/h^2$. The ratio of inertia to that stress is $U{h}^2/(\nu L)$, while lateral viscous derivatives are smaller by $(h/L{)}^2$. The two stated smallness conditions therefore reduce the lateral equations to $\mu u_{zz}=p_x$, $\mu v_{zz}=p_y$. The normal component of balance of linear momentum gives $p_z=0$ at leading order; continuity remains $u_x+v_y+w_z=0$. The no-slip, impermeable rigid plates require $u=v=w=0$ at both plate surfaces.

Integrating with those boundary values givesDepth-integrated continuity has no normal boundary flux, so $\nabla \cdot \bar{u}=0$ and $\boxed{{\nabla }^{2}p=0}$ away from the injection point.

Neglecting surface tension, the fluid free boundary has $p=p_{\mathrm{ext}}$, and its normal velocity is the fluid velocity: $\dot{R}={\bar{u}}_r(R)$. Any constant capillary pressure for a circular interface can instead be included in that spatially uniform boundary pressure. Conservation of injected volume gives $2\pi rh{\bar{u}}_r=Q$. HenceThe logarithmic pressure singularity represents the ideal point source, which in a physical cell is replaced by a finite inlet. The formula is an exact radial solution of the depth-averaged model for either sign of $Q$, while $R^2>0$.

Its robustness differs sharply between the signs. With viscous fluid inside a much less viscous exterior, injection displaces the less viscous phase and is stabilizing; suction draws the exterior into the viscous region and is prone to fingering. One can see this directly without relying on a slogan. Write the boundary as $R+a_n\cos n\theta$ and its regular pressure perturbation as $b_n{r}^n\cos n\theta$. Linearizing the constant-pressure condition gives $\delta p(R)=-a_n{p}_0^{\prime }(R)=a_{n}Q/(2\pi hKR)$. The perturbation radial velocity is then $-n{a}_{n}Q/(2\pi h{R}^2)$, and evaluation of the base velocity at the displaced boundary adds $-a_{n}Q/(2\pi h{R}^2)$. ThusModes decay for injection and grow for suction. Accordingly the suction circle is a mathematically possible symmetric solution but generally an unstable prediction, failing before its formal collapse in a perturbed experiment. This is Stability of an interior Hele-Shaw blob under injection and suction.

Let the thin transverse length be $R{e}^{-1/2}$ times the longitudinal scale and write $Y=R{e}^{1/2}y$, $V=R{e}^{1/2}v$. Continuity then remains $u_x+V_Y=0$. Transverse momentum gives $p_Y=0$ to leading order; the quiescent exterior has constant pressure, hence $p_x=0$. Axial viscous diffusion in $x$ is smaller by $R{e}^{-1}$, leavingSymmetry about the jet centreline requires $V=0$ and $u_Y=0$ at $Y=0$. At large positive $Y$, require $u\to 0$ and $u_Y\to 0$. The transverse velocity can approach a finite nonzero scaled entrainment velocity; imposing $V\to 0$ as well would rule out the jet. The physical transverse velocity remains smaller by $R{e}^{-1/2}$ and matches an entraining outer flow.

Using continuity, the momentum equation is ${\partial }_x(u^2)+{\partial }_Y(uV-u_Y)=0$. Integrating across the jet, the boundary terms vanish because $u,u_Y\to 0$ while $V$ is finite. ThusApart from the dimensional density and velocity/length factors, this is the conserved axial momentum flux supplied by the nozzle. Entrainment increases volume flux without adding axial momentum from the resting exterior.

For $u=x^{\alpha }{f}^{\prime }(\eta )$ with $\eta =Y/x^{\beta }$, use the stream function $\psi =x^{\alpha +\beta }f(\eta )$ so that $u={\psi }_Y$ and $V=-{\psi }_x$. Momentum conservation requires $2\alpha +\beta =0$. Inertia and diffusion scale as $x^{2\alpha -1}$ and $x^{\alpha -2\beta }$, so $\alpha +2\beta =1$. Therefore $\boxed{\alpha =-1/3,\beta =2/3}$. In particularSubstitution reduces momentum to $3f^{(3)}+f{f}^{\prime \prime }+(f^{\prime }{)}^2=0$, with $f(0)=f^{\prime \prime }(0)=0$ and $f^{\prime }(\infty )=0$. Integrating first gives $3f^{\prime \prime }+f{f}^{\prime }=0$, the integration constant vanishing at the centreline. A second integration gives $3f^{\prime }+f^2/2=C$. Parameterize its positive constant as $C=18A^2$ and solve the resulting Riccati equation with $f(0)=0$:The momentum integral fixes its scale, since ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{4}zdz=4/3$:The upper-half entrainment is $V\to -2A{x}^{-2/3}$; the negative side has the opposite sign, drawing ambient fluid inward. This is the Bickley jet, valid in the slender far-downstream laminar regime, rather than in the immediate nozzle region.

Let $t(n)$ denote traction: the force per unit area exerted across an oriented internal surface with unit normal $n$. Define ${\sigma }_{ij}$ as component $i$ of the traction on a coordinate plane whose normal is $e_j$. Balance of linear momentum on a shrinking tetrahedron makes the surface-force terms, of order area, dominate its inertia and body forces, of order volume. Resolving its oblique face into the three coordinate projected areas gives Cauchy's traction formulaOpposite faces have opposite tractions, as follows independently from a shrinking pillbox. This fixes both the index and normal conventions.

For a material volume $V(t)$, the Reynolds transport theorem says $d{\int }_{V(t)}fdV/dt={\int }_{V(t)}[f_t+\nabla \cdot (fu)]dV$. Apply it componentwise to $f=\rho u_i$ and use mass conservation. With no body force in this calculation, balance of linear momentum and the divergence theorem giveArbitrariness of the material volume yields $\boxed{\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}}$, with $D/Dt={\partial }_t+u_j{\partial }_j$.

For constant dynamic viscosity and incompressible velocity, ${\partial }_j{u}_j=0$, soThus $\boxed{\rho (u_t+u\cdot \nabla u)=-\nabla p+\mu {\nabla }^{2}u}$.

In fully developed steady pipe flow $u=w(x,y)e_z$, the convective acceleration vanishes. The transverse equations give $p_x=p_y=0$ and the axial equation gives $p_z=\mu (w_{xx}+w_{yy})$. One side depends only on $z$ and the other only on $(x,y)$, so both equal a constant. HenceIf $n$ is outward from the fluid cross-section, the wall-on-fluid axial shear per length is $\mu \oint {\partial }_{n}wds=\mu {\int }_A\Delta wdA=A{p}_z$. The fluid-on-pipe force is its negative:For forward pressure-driven flow $p_z<0$, this force points along the flow. Its value depends on cross-sectional area and pressure drop, even though the detailed velocity profile depends on the shape.

The initial value can first be subtracted: it is ${\mathcal{F}}_0$-measurable and has the same integrability under both measures, since the density has conditional expectation one on ${\mathcal{F}}_0$. Put $N=M-⟨M,X⟩$. Since $dZ=ZdX$, the Itô product rule and the quadratic covariation identity $d⟨M,Z⟩=Zd⟨M,X⟩$ giveThis is a local P-martingale. Stop $N$ at increasing bounded levels; the resulting products with $Z$ are true martingales, since they are local martingales dominated at stopping times by a constant times the uniformly integrable density process. Part (c) then shows that the stopped $N$ processes are Q-martingales. ConsequentlyThe unqualified true-martingale assertion printed in this part needs an additional integrability hypothesis. One sufficient version is a square-integrable P-martingale $M$. Indeed $Z_t\in L^2(P)$ by boundedness of $Y$, and the covariation bound givesThe Doob L2 maximal inequality, ${\mathbb{E}}_P⟨M{⟩}_t<\infty$, and Cauchy-Schwarz inequality therefore give ${\mathbb{E}}_Q{\sup }_{s\le t}|N_s|<\infty$. Its local martingale identities pass to the limit, making it a true Q-martingale. This is Girsanov compensation with square-integrable input.

Here is a counterexample to the broader printed assertion. On $[0,1]$, take $X_s=W_s$, so $Y_s\equiv 1$ and $Z_s=e^{W_s-s/2}$. Put $a=1/2$ andUnder P, $W_a$ has density ${\pi }^{-1/2}{e}^{-x^2}$, so ${\mathbb{E}}_{P}H<\infty$. Independence of future Brownian increments proves $M$ is a continuous true P-martingale. Under Q, $W_s^Q=W_s-s$ is Brownian and the density of $W_a$ is multiplied by $e^{x-1/4}$, so ${\mathbb{E}}_{Q}H=\infty$. The compensated process after $a$ isThe future Q-Brownian increment is independent of $H$ and has positive finite expected absolute value. Thus ${\mathbb{E}}_Q|N_s|=\infty$ for every $s>a$, so it cannot be a true martingale. This proves that Bounded Girsanov drift does not preserve true martingales under compensation.

For the requested application, the Brownian driver does satisfy the needed conditions. To distinguish the state from the density-driving integral, call the state $U$. Set $Y_s=b(U_s)/a(U_s)$; the stated coefficient bounds make this integrand bounded. Applying the local result to $M=W$ givesIt is a continuous local Q-martingale with quadratic variation $s$. The Lévy characterization of Brownian motion therefore makes it a Q-Brownian motion. Substituting $d{W}_s=d{\tilde{W}}_s+[b(U_s)/a(U_s)]ds$ into the original state equation givesThe Brownian driver after the change is $\tilde{W}$; writing it as $W$ again in the last printed equation amounts to renaming that driver.

Boundedness makes each $N_u$ Q-integrable. For $s\le u\le t$, the density formula and conditioning first on ${\mathcal{F}}_u$ giveThe middle equality uses that $N_u$ is ${\mathcal{F}}_u$-measurable and $Z$ is a martingale; the next uses the assumed martingale property of $NZ$. Thus $N$ is a Q-martingale.

Define $\mathbb{Q}$ on ${\mathcal{F}}_t$ by density $Z_t$. Nonnegativity and countable additivity of the integral make it a measure, and $\mathbb{Q}(\Omega )={\mathbb{E}}_P{Z}_t=1$. Strict positivity of $Z_t$ also makes the two measures equivalent. The Radon-Nikodym derivative identity gives ${\mathbb{E}}_Q[1_{A}Y]={\mathbb{E}}_P[1_{A}Y{Z}_t]$, first for simple variables and then for every Q-integrable $Y$ by approximation.

Write $U={\mathbb{E}}_Q[Y\mid {\mathcal{F}}_s]$. The density on ${\mathcal{F}}_s$ is $Z_s={\mathbb{E}}_P[Z_t\mid {\mathcal{F}}_s]$. For every $A\in {\mathcal{F}}_s$,Moreover $Z_{s}U$ is P-integrable because ${\mathbb{E}}_P[Z_s|U|]={\mathbb{E}}_Q|U|<\infty$. Uniqueness in the definition of conditional expectation provesThis is conditional expectation under a martingale density.

The bounded integrand makes $X_s={\int }_0^s{Y}_{u}d{W}_u$ a square-integrable martingale, with quadratic variationThusNovikov's condition applies to the stochastic exponential $Z_s=\exp (X_s-⟨X{⟩}_s/2)$. Hence $Z$ is a true martingale with $\mathbb{E}{Z}_s=Z_0=1$. It is strictly positive because its exponent is finite almost surely.

For simple predictable integrands, the Itô integral is a continuous square-integrable martingale and the Itô isometry givesApproximate a general square-integrable predictable integrand by simple ones. The resulting integral martingales are Cauchy in terminal $L^2$ norm. Completeness of continuous square-integrable martingales guarantees a unique continuous martingale limit, independent of the approximation, which defines the integral on the completed integrand space.

Work modulo indistinguishability of stochastic processes, with norm $\Vert M\Vert =(\mathbb{E}|M_t{|}^2{)}^{1/2}$. It is a genuine norm: zero terminal value implies $M_s=\mathbb{E}[M_t\mid {\mathcal{F}}_s]=0$ at every fixed time, and continuity extends equality from countably many rational times to all times simultaneously.

Let $M^j$ be a Cauchy sequence in this norm. Choose a subsequence with $\Vert M^{j_{n+1}}-M^{j_n}\Vert <2^{-n}$. Part (b) applies to these difference martingales, so their telescoping series converges uniformly almost surely. Its limit has continuous paths. At each time $s$, conditional-expectation contractivity givesso the fixed-time limit also exists in $L^2$ and agrees with the pathwise limit. It has an adapted version obtained by taking the finite pointwise subsequence limit when it exists, and zero otherwise. Each such definition is measurable at its own time, and the almost surely uniform limit gives continuous paths. Passing to the $L^2$ limit in $\mathbb{E}[M_v^j\mid {\mathcal{F}}_s]=M_s^j$ shows that the limit is a martingale. The original Cauchy sequence converges to it in terminal norm. Thus the space is complete.

Part (b) supplies what fixed-time $L^2$ completeness alone does not: a single full-probability set on which a subsequence converges uniformly in time, preserving path continuity. This is completeness of continuous square-integrable martingales.

Set $A_n=\{{\sup }_{s\le t}|M_s^{(n)}|\ge 2^{-(n-1)/2}\}$. Part (a) gives ${\sum }_n\mathbb{P}(A_n)<\infty$. The Borel-Cantelli first lemma implies that, almost surely, only finitely many $A_n$ occur. On the complementary full-probability event, all sufficiently late terms satisfyThese deterministic bounds form a convergent geometric series. The finitely many earlier suprema are finite by continuity on the compact time interval. HenceThe series therefore converges absolutely and uniformly on $[0,t]$ by the Weierstrass M-test, and its sample-path limit is continuous.

The given Doob L2 maximal inequality givesApply Markov's inequality to the nonnegative variable $(M_t^{\ast }{)}^2$, at the threshold printed in the question:The corresponding threshold for $M_t^{\ast }$ itself is $2^{-(n-1)/2}$.